Question

Assume that you want to construct a voltaic cell that uses the following half reactions: $$ \begin{array}{ll} \mathrm{A}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{A}(s) & E_{\mathrm{red}}^{\circ}=-0.10 \mathrm{~V} \\ \mathrm{~B}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{B}(s) & E_{\mathrm{red}}^{\circ}=-1.10 \mathrm{~V} \end{array} $$ You begin with the incomplete cell pictured below, in which the electrodes are immersed in water. (a) What additions must you make to the cell for it to generate a standard emf? (b) Which electrode functions as the cathode? (c) Which direction do electrons move through the external circuit? (d) What voltage will the cell generate under standard conditions? [Sections \(20.3\) and 20.4]

Short answer

To construct the voltaic cell, add 1M electrolyte solutions containing A²⁺ and B²⁺ cations. The cathode is reaction A with a standard reduction potential of -0.10 V, and the anode is reaction B with -1.10 V. Electron flow occurs from electrode B to electrode A in the external circuit. The standard cell potential is 1.00 V, so under standard conditions, the cell generates a voltage of 1.00 V.

Step-by-step solution

Find the anode and cathode reactions

First, compare the standard reduction potentials of the two half-reactions. The reaction with the higher standard reduction potential (E°) will proceed as a reduction, while the reaction with the lower E° will proceed as an oxidation. In this case, since E°(A) > E°(B), reaction A will be the reduction (cathode) half-reaction and reaction B will be the oxidation (anode) half-reaction.

Determine the required additions

To complete the cell and generate a standard emf, you need to add electrolyte solutions containing the cations corresponding to each half-reaction, i.e., A²⁺ and B²⁺. These solutions should be of 1M concentration for a standard cell.

Calculate standard cell potential

The standard cell potential (E°ᶜᵉˡˡ) is calculated based on the standard reduction potentials of the anode and cathode half-reactions: E°ᶜᵉˡˡ = E°(cathode) - E°(anode) In this case, the cathode is reaction A, and the anode is reaction B: E°ᶜᵉˡˡ = (-0.10 V) - (-1.10 V) = 1.00 V

Identify the cathode

As determined earlier, reaction A functions as the cathode since it has higher standard reduction potential (E°) and undergoes reduction.

Determine the electron flow direction

In the external circuit, electrons flow from the anode (where oxidation occurs) to the cathode (where reduction occurs). In this case, electrons will move from electrode B (the anode) to electrode A (the cathode).

Compute cell voltage

Under standard conditions, the cell voltage will be equal to the standard cell potential previously calculated. Thus, Vᶜᵉˡˡ = E°ᶜᵉˡˡ = 1.00 V.

Key concepts

Standard Reduction Potential

Understanding the standard reduction potential is crucial when constructing a voltaic cell. It's a measure that indicates the tendency of a chemical species to acquire electrons and be reduced. This measure is essential when we want to foresee which half-reaction will occur at which electrode in the cell.

Each half-reaction has an associated standard reduction potential, typically measured in volts (V) and listed under standard conditions, which means a temperature of 298 K, a 1 M concentration for each ion, and a pressure of 1 atm for any gases involved. The more positive the standard reduction potential, the greater the substance's affinity for electrons. For example, if we have two half-reactions, the one with the higher (less negative) standard reduction potential is naturally the reduction half-reaction that occurs at the cathode.

By looking at the provided half-reactions for A and B:\[\mathrm{A}^{2+}(aq)+2 \mathrm{e}^{-} \longrightarrow \mathrm{A}(s), E_{\mathrm{red}}^{\circ}=-0.10 \mathrm{~V}\] and \[\mathrm{B}^{2+}(aq)+2 \mathrm{e}^{-} \longrightarrow \mathrm{B}(s), E_{\mathrm{red}}^{\circ}=-1.10 \mathrm{~V}\], we observe that A has a less negative standard reduction potential and so it will function as the cathode.

Cathode and Anode Identification

In a voltaic cell, the cathode is the electrode where the reduction takes place, and the anode is where oxidation occurs. Identification of these components is guided by standard reduction potentials of the involved half-reactions. The substance with a higher (or less negative) potential will be reduced and, therefore, make the cathode.

Back to our example, the substance A with standard reduction potential \(E_{\mathrm{red}}^{\circ}=-0.10 \mathrm{~V}\) is less negative compared to B's \(E_{\mathrm{red}}^{\circ}=-1.10 \mathrm{~V}\), which implies that A is the cathode and B is the anode. Remember, the cathode attracts cations because it is relatively more positive and the anode attracts anions.

Electrons always flow from the anode to the cathode during the operation of the cell, meaning for our cell, electron flow will be from electrode B to electrode A. Adding the appropriate electrolyte solutions and ensuring standard conditions are met allows the cell to function correctly and effectively.

Calculation of Cell Potential

The potential difference created by the voltaic cell, referred to as the electromotive force (emf) or cell potential, is calculated by taking the difference between the cathode's and anode's standard reduction potentials. It’s depicted by the equation:\[E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\].

In the context of the given exercise, the calculation would be as follows:\[E^{\circ}_{\text{cell}} = (-0.10\,V) - (-1.10\,V) = 1.00\,V\]. This calculation tells us that the potential difference, or voltage, produced under standard conditions by our voltaic cell, is 1.00 V.

The calculated cell potential helps in determining the feasibility and efficiency of a voltaic cell. It’s a key factor in the design and application of cells for real-world uses such as batteries. A high cell potential generally indicates a more effective cell capable of providing greater energy per unit of charge.